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Solutions: Molarity

Solutions: Molarity. Chemistry Ms. Piela. Solutions: Molarity. I. Concentration of Solutions. A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent. B. Dilute Solution – contains a lower concentration of solute.

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Solutions: Molarity

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  1. Solutions: Molarity Chemistry Ms. Piela

  2. Solutions: Molarity • I. Concentration of Solutions • A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent • B. Dilute Solution – contains a lower concentration of solute • C. Concentrated Solution – contains a higher concentration of solute

  3. Solutions: Molarity • II Molarity • The number of moles of solute dissolved per liter (L) of solution • B. Sometimes referred to as a molarsolution (i.e. 0.5 molar solution • A. The most important unit of concentration in chemistry • C. To determine molarity, DIVIDEthe number of MOLESby the volume in LITERS

  4. Solutions: Molarity • II Molarity D. Example Problems • What is the molarity of a solution that contains 2.0 mol of glucose in 5.0 L of water? 0.4 M

  5. Solutions: Molarity • II Molarity D. Example Problems • A saline solution contains 0.90 g NaCl in exactly 0.100 L of solution. What is the molarity of the solution? 0.20L

  6. Solutions: Molarity • II Molarity D. Example Problems • How many moles of solute are present in 1.5 L of 0.24 M Na2SO4? 0.36 mol

  7. Solution Stoichiometry Stoichiometry calculations can be done using molarity and volumes. Since stoichiometry just depends on mole ratios, the goal is to find moles!

  8. Example Problem #1 Calculate the number of mL of 2.00 M HNO3 solution required to react with 216 grams of Ag according to the equation. 3 Ag(s) + 4 HNO3 (aq)→ 3 AgNO3 (aq) + NO(g) + 2 H2O(l)

  9. Example Problem #2 Calculate in mL the volume of 0.500 M NaOH required to react with 3.0 grams of acetic acid (HC2H3O2). The equation is: NaOH(aq) + HC2H3O2 (aq)→ NaC2H3O2 (aq) + H2O(l)

  10. Example Problem #3 Calculate the number of grams of AgCl formed when 0.200 L of 0.200 M AgNO3 reacts with an excess of CaCl2. The equation is: 2 AgNO3 (aq) + CaCl2 (aq)→ 2 AgCl(s) + Ca(NO3)2 (aq)

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